- #1

Juntao

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Okay, this is probably a simple problem, but I don't know how to start it. Help appreciated.

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- Thread starter Juntao
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- #1

Juntao

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- 0

Okay, this is probably a simple problem, but I don't know how to start it. Help appreciated.

- #2

Hurkyl

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What do you get when you plug what you know into that formula? (represent unknowns with variables)

- #3

Juntao

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Um, I think this is the one, hopefully:

The derivative of f at Po in the direction of u= gradient f dot with u = magnitude grad f * magnitude u * cos theta = magnitude grad f * cos theta.

Or like D_u f= grad f (dot) u = |grad f| |u| cos theta = |grad f| cos theta

Maybe that'll help me out?

- #4

HallsofIvy

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Yes, that's true as long as u is a unit vector (has length 1).D_u f= grad f (dot) u

Another way of saying this is that the directional derivative is

f_x u_x+ f_y u_y where f_x is the partial derivative and u_x is the x component of unit vector u.

Find unit vectors in both the given directions and write out

f_x u_x+ f_y u_y= √(2) and

f_x v_x+ f_y v_y= √(5).

That gives you two equations for f_x and f_y.

- #5

Hurkyl

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D_u f= grad f (dot) u

That's the one of which I was thinking.

You know two instances of

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